Let $G$ be a connected algebraic group of positive dimension. Prove that $G$ is semisimple if and only if $G$ has no closed connected commutative normal subgroup except $e$.

It is clear in one direction, given the definition of semisimple algebraic groups. In the other direction, suppose $G$ is not semisimple, and let $H=R(G) \neq \{e\}$, the radical of $G$. As $H$ is solvable, it has a lower central series:

Hence, $H_n$ is commutative and $\neq \{e\}$. The contradiction can be found if $H_n \unlhd G$. But in fact, I have only $H \unlhd G$. Normality is in general not liftable. So, I don't know what to do. Maybe other nontrivial closed connected commutative normal subgroup can be found?